There is no information in the cage b1-a2-b2, but let’s solve it quickly:

We will use the cage “1:” in d4-e4-e5 as the “key”. This cage has five possible combinations 1-2-2, 1-3-3, 1-4-4, 1-5-5 and 2-2-4. The cage “60x” in c4-c5-d5 has the only combination 5-3-4 so 1-3-3, 1-4-4 and 1-5-5 are not valid in “1:”. The cage “60x” (60 = 2 x 2 x 3 x 5) in e1-e2-d3-e3 has two possible combinations, 5-3-2-2 and 5-3-4-1, but 5-3-2-2 is not valid with any of the other two combinations for “1:” (1-2-2 or 2-2-4).

I have shown in blue the hypothesis 2-2-4. Now the 4 in e4 forces the 4 of 5-3-4-1 to go to d3 and this is the relevant point: looking to columns 3 and 4 we have 12 + 2 + 4 + 10 = 28, leaving a total of 30 (2 columns) - 28 = 2 for the sum of c2 and c3, which is impossible.

Once we have determined the 1-2-2 (now in black) as the final solution for “1:”, the 1 of 5-3-4-1 goes to d3 and that means that 30 - (12 + 2 + 1 +10) = 5, that is, the sum of c2-c3 is 5. As we have a 4 in b4, b3 must be 3 (12 - 4 - 5) (shown in green). Since c2-c3 have a sum of 5 and 2-3 is not possible now, they must be 1-4 so c2 = 1 and c3 = 4 (and c5 = 3, d5 = 4). The rest is very easy (completed in orange).

Then, we have solved the entire puzzle with some analysis but regardless of the info of the cage b1-a2-b2 (it is actually “2:”, but it could have been “8x”, “1-“ or “7+”, it is indifferent). And, certainly, we have not needed any additional info, provided as a help, from the twin puzzle (this is the 5x5 difficult on Wednesday Oct 05, 2011, Puzzle id: 361005):

Now let’s see these puzzles (all of them with a unique solution):

We do not need any other information to solve them, and in fact they take very little time (I provide the solutions at the end but try to solve them, you will only need a few seconds). In the case of the 4x4, 6 cells (among 16) have “information”, that is, the 37,50% of the surface of the puzzle. In the case of the 5x5, 9 cells (among 25) have “information”, that is, the 36,00%, and in the case of the 6x6 (12 out of 36) the info represents the 33,33%. Even a 3x3 puzzle with, i.e., a cage “1-“ in a2-a3-b3 (that would be enough to solve the puzzle), has a 33,33% of “info”. Certainly these are very particular puzzles but the question is: How much is the “essential” information in a puzzle to arrive to the unique solution?. How much information is “superfluous”? Is it always possible to remove information leaving generously “blank” cages (like a donation to the program, like when in chess some initial advantage is given to the opponent donating the whites, a pawn or even a knight)?. How can we build a puzzle in which the 100,00% of the information is necessary having, of course, a reasonable size of the cages, and not a extremely reduced number of them?.